To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.
If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.
![F = ma \rightarrow a = \frac{F}{m}](https://tex.z-dn.net/?f=F%20%3D%20ma%20%5Crightarrow%20a%20%3D%20%5Cfrac%7BF%7D%7Bm%7D)
Replacing,
![a =\frac{42N}{83000kg}](https://tex.z-dn.net/?f=a%20%3D%5Cfrac%7B42N%7D%7B83000kg%7D)
![a =5.06*10^{-4}m/s^2](https://tex.z-dn.net/?f=a%20%3D5.06%2A10%5E%7B-4%7Dm%2Fs%5E2)
The total speed change
we have that the value is 0.71m/s
If we know that acceleration is the change of speed in a fraction of time,
![a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7B%5CDelta%20v%7D%7Bt%7D%20%5Crightarrow%20t%20%3D%20%5Cfrac%7B%5CDelta%20v%7D%7Ba%7D)
We have that,
![t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B0.71m%2Fs%7D%7B5.06%2A10%5E%7B-4%7Dm%2Fs%5E2%20%7D)
![t = 1403.16s](https://tex.z-dn.net/?f=t%20%3D%201403.16s)
Therefore the Rocket should be fired around to 1403.16s
Answer:
Option D, The equator gets more direct sunlight throughout the yea
Explanation:
question: Please help!!!
If a bottle is being squeezed with a force of 10 Newtons and the area of the bottle is 15
squared inches. What is the pressure??
Answer:
1025.64 N/m²
Explanation:
Pressure: This can be defined as the ratio of force to area. The si unit of pressure is N/m².
From the question,
P = F/A........................ Equation 1
Where F = Force, A = Area.
Given: F = 10 Newtons, A = 15 Squared Inches = (15×0.00065) = 0.00975 m²
Substitute these values into equation 1
P = 10/0.00975
P = 1025.64 N/m²
Hence the pressure of the bottle is 1025.64 N/m²
Answer:
1.61 second
Explanation:
Angle of projection, θ = 53°
maximum height, H = 7.8 m
Let T be the time taken by the ball to travel into air. It is called time of flight.
Let u be the velocity of projection.
The formula for maximum height is given by
![H = \frac{u^{2}Sin^{2}\theta }{2g}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7Bu%5E%7B2%7DSin%5E%7B2%7D%5Ctheta%20%7D%7B2g%7D)
By substituting the values, we get
![7.8= \frac{u^{2}Sin^{2}53 }{2\times 9.8}](https://tex.z-dn.net/?f=7.8%3D%20%5Cfrac%7Bu%5E%7B2%7DSin%5E%7B2%7D53%20%7D%7B2%5Ctimes%209.8%7D)
u = 9.88 m/s
Use the formula for time of flight
![T = \frac{2uSin\theta }{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2uSin%5Ctheta%20%7D%7Bg%7D)
![T = \frac{2\times 9.88\times Sin53 }{9.8}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Ctimes%209.88%5Ctimes%20Sin53%20%7D%7B9.8%7D)
T = 1.61 second
When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling
the capacitance of the capacitor in air, the charge Q, the capacitance
and the voltage (
) are related by
(1)
when the source is disconnected the charge Q remains on the capacitor.
When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):
![C=k C_0 = 5.4 C_0](https://tex.z-dn.net/?f=%20C%3Dk%20C_0%20%3D%205.4%20C_0%20)
this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:
![V=\frac{Q}{C}=\frac{Q}{5.4 C_0}](https://tex.z-dn.net/?f=%20V%3D%5Cfrac%7BQ%7D%7BC%7D%3D%5Cfrac%7BQ%7D%7B5.4%20C_0%7D%20%20%20)
And since
, substituting into the previous equation, we find:
![V=\frac{C_0 V_0}{5.4 C_0}=\frac{V_0}{5.4}=\frac{391 V}{5.4}=72.4 V](https://tex.z-dn.net/?f=%20V%3D%5Cfrac%7BC_0%20V_0%7D%7B5.4%20C_0%7D%3D%5Cfrac%7BV_0%7D%7B5.4%7D%3D%5Cfrac%7B391%20V%7D%7B5.4%7D%3D72.4%20V%20%20%20%20)